`2.68xx10^(-3)` moles of solution containing anion `A^(n+)` require `1.61xx10^(-3)` moles of `MnO_(4)^(-)` for oxidation of `A^(n+)` to `AO_(3)^(-)` in acidic medium. What is the value of `n`?

Explanation

To solve the problem, we need to determine the value of \( n \) in the anion \( A^{n+} \) based on the given moles of \( A^{n+} \) and \( MnO_4^{-} \) required for the oxidation reaction.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Moles of \( A^{n+} = 2.68 \times 10^{-3} \) moles
- Moles of \( MnO_4^{-} = 1.61 \times 10^{-3} \) moles

2. **Determine the Change in Oxidation State:**
- The oxidation state of manganese in \( MnO_4^{-} \) in acidic medium is +7, which is reduced to +2. This means that \( MnO_4^{-} \) gains 5 electrons during the reaction.
- The oxidation of \( A^{n+} \) to \( AO_3^{-} \) involves a change in oxidation state. Let’s denote the oxidation state of \( A \) in \( AO_3^{-} \) as \( x \).
- The equation for the oxidation state of \( A \) in \( AO_3^{-} \) is:
\[ x - 2 \times 3 = -1 \implies x - 6 = -1 \implies x = +5 \]
- Therefore, the oxidation state of \( A \) changes from \( n \) to +5, indicating that \( A^{n+} \) loses \( (5 - n) \) electrons during oxidation.

3. **Calculate the Equivalents:**
- The equivalents of a substance in a redox reaction can be calculated using the formula:
\[ \text{Equivalents} = \text{Moles} \times \text{Valency Factor} \]
- For \( MnO_4^{-} \):
\[ \text{Equivalents of } MnO_4^{-} = 1.61 \times 10^{-3} \text{ moles} \times 5 \text{ (valency factor)} \]
- For \( A^{n+} \):
\[ \text{Equivalents of } A^{n+} = 2.68 \times 10^{-3} \text{ moles} \times (5 - n) \]

4. **Set the Equivalents Equal:**
- Since the equivalents of \( MnO_4^{-} \) and \( A^{n+} \) are equal, we can set up the equation:
\[ 1.61 \times 10^{-3} \times 5 = 2.68 \times 10^{-3} \times (5 - n) \]

5. **Solve for \( n \):**
- Calculate the left side:
\[ 1.61 \times 10^{-3} \times 5 = 8.05 \times 10^{-3} \]
- Now, substitute this into the equation:
\[ 8.05 \times 10^{-3} = 2.68 \times 10^{-3} \times (5 - n) \]
- Divide both sides by \( 2.68 \times 10^{-3} \):
\[ \frac{8.05 \times 10^{-3}}{2.68 \times 10^{-3}} = 5 - n \]
- Calculate the left side:
\[ 3 = 5 - n \]
- Rearranging gives:
\[ n = 5 - 3 = 2 \]

### Final Answer:
The value of \( n \) is **2**.